find ⌊∫_0 ^(2023) (2/(x+e^x ))dx⌋=?


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Question Number 204409 by mr W last updated on 16/Feb/24

$f i n d ⌊ \int_{0}^{2023} \frac{2}{x + e^{x}} d x ⌋ = ?$
Commented by witcher3 last updated on 16/Feb/24

$nice problems sir Thanx for share it$
	Answered by witcher3 last updated on 16/Feb/24

	$⌊ x ⌋ floor function ?$
	Commented by mr W last updated on 16/Feb/24

	$y e s .$
	Answered by mr W last updated on 18/Feb/24

	$x \in [0, 2023] \Rightarrow x ⩾ 0$ $\frac{2}{x + e^{x}} < \frac{2}{e^{x}}$ $\Rightarrow \int_{0}^{2023} \frac{2}{x + e^{x}} d x < \int_{0}^{2023} \frac{2}{e^{x}} d x = 2 {[- \frac{1}{e^{x}}]}_{0}^{2023} = 2 (1 - \frac{1}{e^{2023}}) < 2$
	Commented by mr W last updated on 18/Feb/24

	Commented by mr W last updated on 18/Feb/24

	$y = \frac{2}{x + e^{x}}$ $y^{'} = - \frac{2 (1 + e^{x})}{{(x + e^{x})}^{2}} < 0$ $\Rightarrow y i s d e c r e a s i n g$ $y^{″} = - \frac{2 e^{x}}{{(x + e^{x})}^{2}} + \frac{4 {(1 + e^{x})}^{2}}{{(x + e^{x})}^{3}} = \frac{2 [2 + 4 e^{x} + e^{x} (e^{x} - x)]}{{(x + e^{x})}^{3}} > 0$ $\Rightarrow y i s ⌣$ $\int_{0}^{2023} y d x = r e d a r e a > b l u e a r e a$ $. . .$
	Answered by witcher3 last updated on 16/Feb/24

	$⌊ x + y] ⩾ [x] + [y]$ $[\int_{0}^{2023} \frac{2 dx}{x + e^{x}}] = ⌊ \underset{k = 0}{\sum^{2022}} \int_{k}^{k + 1} \frac{2 dx}{x + e^{x}} ⌋ ⩽ \underset{k = 0}{\sum^{2022}} [\int_{k}^{k + 1} \frac{2 dx}{x + e^{x}}]$ $\underset{k = 0}{\sum^{2022}} ⌊ \int_{k}^{k + 1} \frac{2}{x + e^{x}} dx ⌋ = [\int_{0}^{1} \frac{2 dx}{x + e^{x}}] + \underset{k = 1}{\sum^{2022}} [\int_{k}^{k + 1} \frac{2 dx}{x + e^{x}}]$ $\int_{k}^{k + 1} \frac{2 dx}{x + e^{x}} < \int_{k}^{k + 1} \frac{2 dx}{k + e^{k}} = \frac{2}{k + e^{k}} < 1; \forall k ⩾ 1; ‘ ‘ e^{1} > 2^{″}$ $\Rightarrow \underset{k = 1}{\sum^{2022}} [\int_{k}^{k + 1} \frac{2 dx}{x + e^{x}}] = 0$ $\int_{0}^{1} \frac{2 dx}{x + e^{x}}$ $e^{x} ⩾ 1 + x \Rightarrow \frac{1}{e^{x} + x} ⩽ \frac{1}{2 x + 1} \Rightarrow \int_{0}^{1} \frac{2 dx}{x + e^{x}} ⩽ \int_{0}^{1} \frac{2 dx}{2 x + 1} = {[\ln (2 x + 1)]}_{0}^{1} = \ln (3)$ $\Rightarrow [\int_{0}^{2023} \frac{2 dx}{x + e^{x}}] ⩽ [\ln (3)] = 1$ $\Rightarrow [\int_{0}^{2023} \frac{2 dx}{x + e^{x}}] ⩽ 1$ $\int_{0}^{1} \frac{2 dx}{x + e^{x}} ⩾ \int_{0}^{1} 2 \frac{dx}{1 + e^{x}} = 2 \ln (\frac{2}{1 + e^{- 1}})$ $\int_{1}^{2} \frac{2 dx}{x + e^{x}} ⩾ \int_{1}^{2} \frac{2 dx}{2 + e^{x}} = \ln (\frac{1 + {2 e}^{- 1}}{1 + {2 e}^{- 2}})$ $\int_{0}^{2} \frac{2 dx}{x + e^{x}} ⩾ \ln (\frac{4 (1 + {2 e}^{- 1})}{{(1 + e^{- 1})}^{2} (1 + {2 e}^{- 2})})$ $4 (1 + {2 e}^{- 1}) ⩾ e (1 + {2 e}^{- 1} + e^{- 2}) (1 + {2 e}^{- 2})$ $4 + {8 e}^{- 1} ⩾ {2 e}^{- 1} + {4 e}^{- 2} + {2 e}^{- 3} + e + 2 + e^{- 1}$ $2 + {5 e}^{- 1} ⩾ {4 e}^{- 2} + {2 e}^{- 3} + e$ $e^{4} + 2 + 4 e - {2 e}^{3} - {5 e}^{2} ⩽ 0$ $e^{3} (e - 2 - {3 e}^{- 1}) + 2 + 4 e - {2 e}^{2} < 0 . . . True$ ${3 e}^{- 1} > 1 \Rightarrow e - 2 - {3 e}^{- 1} < e - 3 < 0$ $2 + 4 e - {2 e}^{2} = 2 (1 + 2 e - e^{2});$ $Tackx \overset{p}{\to} 1 + 2 x - x^{2}$ $p^{'} (x) = 2 - 2 x$ $x ⩾ 1 decreade e > 2.5 \Rightarrow p (e) < p (2.5) = 1 + 5 - {(2.5)}^{2} = - 0.25 < 0$ $\Rightarrow \ln (\frac{4 (1 + {2 e}^{- 1})}{{(1 + e^{- 1})}^{2} (1 + {2 e}^{- 2})}) ⩾ \ln (e) = 1$ $\int_{0}^{2023} \frac{2 dx}{x + e^{x}} ⩾ \int_{0}^{2} \frac{2 dx}{x + e^{x}} ⩾ 1$ $\Rightarrow ⌊ \int_{0}^{2023} \frac{2 dx}{x + e^{x}} ⌋ ⩾ 1 \Rightarrow enswer = 1$
	Commented by mr W last updated on 17/Feb/24

	$r i g h t a n s w e r! t h a n k s!$
	Commented by witcher3 last updated on 17/Feb/24

	$withe pleasur$
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